Question 14.5: For an equivalent noise bandwidth of 10 MHz and a total nois......
For an equivalent noise bandwidth of 10 MHz and a total noise power of 0.0276 pW, determine the noise density and equivalent noise temperature.
Substituting into Equation 12, we have
T_{e}= T\!(F-1) (12)
N_{0}=\frac{N}{B}=\frac{276\,×\,10^{-16}\, W}{10\,×\,10^{6} \,Hz}=276\, ×\, 10^{-23}\, W/Hz
or simply 276 × 10^{-23} W.
N_{0}=10\, \log(276 × 10^{-23})= -205.6 dBW/Hz
or simply −205.6 dBW. Substituting into Equation 13 gives us
T_{e(dBK)} = 10\, \log\, T_{e} (13)
N_{0}=10\, \log\, 276 × 10^{-16} −10 \,\log \,10\,MHZ
= -135.6 dBW − 70 dB = −205.6 dBW
Rearranging Equation 12 and solving for equivalent noise temperature yields
T_{e}=\frac{N_{0}}{K}
=\frac{276\,×\,10^{-23}\, J/cycle}{1.38\,×\,10^{-23}\, J/K}= 200 \,K/cycle
Expressed as a log, T_{e}= 10 log 200 = 23 dBK
=N_{0} \,-\, 10\, \log\, K = N_{0} \,-\, 10\, \log\, 1.38 × 10^{-23}
= −205.6 dBW − (−228.6 dBWK) = 23 dBK